162 7983 83126carbonsand6x1272so83 7211c6h11br so262

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Unformatted text preview: because there is no –OH peak present. Acetopheone can be eliminated next because there is no peak for aromatic C-H stretches. Also the carbonyl peak would be at a higher energy like around 1800 cm-1. 2-methyl-2-cyclohexenone can be eliminated because there is no O carbon carbon double bond peak. Leaving 4-ethylcyclohexanone. (4) Determine the structure of the compound that gives rise to the following mass and IR spectra. The molecular ion peak is at 162 and the M+2 peak is at 164, and they are in a 1:1 ratio, therefore there is a bromine atom. 162-79=83 83/12=6 carbons and 6x12= 72 so 83-72=11 C6H11Br So 2(6)+2-11-1=2/2=1, which means 1 double bond or 1 ring. Looking at the IR, there is no C=C peak so that means a ring. Br (5) Determine the structure of the compound that gives rise to the following mass and IR spectra. So the molecular ion peak is 72. So 72/12= 6 carbons 2(6)+2=14/2=7 a bit high. So subtract 1 C and replace with 12 Hs. C5H12. 2(5)+2-12=0 Not pentane, there is a carbony...
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This note was uploaded on 02/07/2014 for the course CHEM 2311 taught by Professor Stenberg during the Spring '11 term at Northeastern.

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